Power in Balanced 3 Phase Loads

Balanced loads, in a 3φ system, have identical impedance in each secondary winding (Figure 12). The impedance of each winding in a delta load is shown as Z_∆ (Figure 12a), and the impedence in a wye load is shown as Z_y (Figure 12b). For either the delta or wye connection, the lines A, B, and C supply a 3φ system of voltages.

Figure 12 : 3φ Balanced Loads

In a balanced delta load, the line voltage (V_L) is equal to the phase voltage (V_ø), and the line current (I_L) is equal to the square root of three times the phase current (√3I_ø).

The below Equation is a mathematical representation of V_L in a balanced delta load.

V_L = V_ø

The below Equation is a mathematical representation of I_L in a balanced delta load.

I_{L =}√3I_ø

In a balanced wye load, the line voltage (V_L) is equal to the square root of three times phase voltage (√3V_ø), and line current (I_L) is equal to the phase current (I_ø).

The below Equation is a mathematical representation of V_L in a balanced wye load.

V_L = √3V_ø

The below Equation is a mathematical representation of I_L in a balanced wye load.

I_{L =} I_ø

Because the impedance of each phase of a balanced delta or wye load has equal current, phase power is one third of the total power.

The below Equation is the mathematical representation for phase power (P_ø) in a balanced delta or wye load.

P_ø = V_ø I_ø cosθ

Total power (P_T) is equal to three times the single-phase power.

The below Equation is the mathematical representation for total power in a balanced delta or wye load.

P_T = 3 V_ø I_ø cosθ

In a delta-connected load,

In a wye-connected load,

As you can see, the total power formulas for delta- and wye-connected loads are identical.

Total apparent power (S_T) in volt-amperes and total reactive power (Q_T) in volt-amperes-reactive are related to total real power (P_T) in watts (Figure 13).

Figure 13 : 3φ Power Triangle

A balanced three-phase load has the real, apparent, and reactive powers given by:

Example 1:

Each phase of a delta- connected 3φ AC generator supplies a full load current of 200 A at 440 volts with a 0.6 lagging power factor, as shown in Figure 14.

Figure 14 : Three-Phase Delta Generator

Find:

1. V_L
2. I_L
3. P_T
4. Q_T
5. S_T

Solution :

1. Calculate V_L

V_L = V_ø

V_L = 440V

2. Calculate I_L

I_{L =}√3I_ø

I_L  = 1.73 x 200

I_L  = 346 amps

3. Calculate P_T

P_T = √3 V_L I_L cosθ

P_T = 1.73 x 440 x 346 x 0.6

P_T = 158.2 kW

4. Calculate Q_T

Q_T = √3 V_L I_L sinθ

Q_T = 1.73 x 440 x 346 x 0.8

Q_T = 210.7 KVR

5. Calculate S_T

S_T = √3 V_L I_L

S_T = 263.4 KVA

Example 2:

Each phase of a wye- connected 3φ AC generator supplies a 100 A current at a phase voltage of 240V and a power factor of 0.9 lagging, as shown in Figure 15.

Figure 15 : Three-Phase Wye Generator

Find:

1. V_L
2. P_T
3. Q_T
4. S_T

Solution :

1. Calculate V_L

V_L = √3V_ø

V_L = 1.73 x 240

V_L = 415.2 volts

2. Calculate P_T

P_T = √3 V_L I_L cosθ

P_T = 1.73 x 415.2 x 100 x 0.9

P_T = 64.6 kW

3. Calculate Q_T

Q_T = √3 V_L I_L sinθ

Q_T = 1.73 x 415.2 x 100 x 0.436

Q_T = 31.3 KVAR

4. Calculate S_T

S_T = √3 V_L I_L

S_T = 1.73 x 415.2 x 100

S_T = 71.8 KVA